A resistor R and $2 μ F$ capacitor in series is connected through a switch to $200 V$ direct supply. Across the capacitor is a neon bulb that lights up at $120 V$ . The value of R to make the bulb light up 5 s after the switch has been closed is $x \times 10^{6} Ω$ then x is nearest to which integer $(\log_{10} 2.5 = 0.4)$

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Detailed Solution

$\begin{array}{l} V_{R} = i R \\ V_{R} = i_{o} [e^{\frac{- t}{C R}}] R \\ V_{R} = V_{0} [e^{\frac{- t}{C R}}] \end{array}$

$V_{c} = 200 (1 - e^{- t / R C})$

$120 = 200 (1 - e^{- t / R C})$

$e^{- t / R C} = \frac{200 - 120}{200} = \frac{80}{200}$

$+ \frac{t}{R C} = \log_{e} (2.5) = \frac{0.4}{0.43} = 0.92$

$\Rightarrow R = \frac{5}{0.92 \times 2 \times 10^{- 6}}$

$\Rightarrow R = 2.71 \times 10^{6} Ω$

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